The provided sample mean is
"\\bar X = 8.93" and the known population standard deviation is σ=1.4, and the sample size is n = 30.
The following null and alternative hypotheses need to be tested:
"\\\\\n\nH_0:\u03bc\u22659.72\\\\\n\nHa:\u03bc<9.72"
This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.
The z-statistic is computed as follows:
"z=\\dfrac{\\overline{x}-\\mu_0}{\\frac{\\sigma}{\\sqrt{n}}}"
"z=\\dfrac{8.93-9.72}{\\frac{1.4}{\\sqrt{30}}}" = −3.091
Using Z table, are
The p-value is p = 0.001, P(Z<=-3.091)= 0.001,since p=0.001<0.05, it is concluded that the null hypothesis is rejected.
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 9.72, at the 0.05 significance level.
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